Uniqueness of spaces pretangent to metric spaces at infinity
We find the necessary and sufficient conditions under which an unbounded metric space X has, at infinity, a unique pretangent space Ωˣ∞, ř for every scaling sequence ř. In particular, it is proved that Ωˣ∞, ř is unique and isometric to the closure of X for every logarithmic spiral X and every ř. It...
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Інститут прикладної математики і механіки НАН України
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nasplib_isofts_kiev_ua-123456789-1694322025-02-09T20:35:53Z Uniqueness of spaces pretangent to metric spaces at infinity Dovgoshey, O. Bilet, V. We find the necessary and sufficient conditions under which an unbounded metric space X has, at infinity, a unique pretangent space Ωˣ∞, ř for every scaling sequence ř. In particular, it is proved that Ωˣ∞, ř is unique and isometric to the closure of X for every logarithmic spiral X and every ř. It is also shown that the uniqueness of pretangent spaces to subsets of a real line is closely related to the “asymptotic asymmetry” of these subsets. 2019 Article Uniqueness of spaces pretangent to metric spaces at infinity / O. Dovgoshey, V. Bilet // Український математичний вісник. — 2019. — Т. 16, № 1. — С. 57-87. — Бібліогр.: 12 назв. — англ. 1810-3200 2010 MSC. 54E35 https://nasplib.isofts.kiev.ua/handle/123456789/169432 ru Український математичний вісник application/pdf Інститут прикладної математики і механіки НАН України |
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We find the necessary and sufficient conditions under which an unbounded metric space X has, at infinity, a unique pretangent space Ωˣ∞, ř for every scaling sequence ř. In particular, it is proved that Ωˣ∞, ř is unique and isometric to the closure of X for every logarithmic spiral X and every ř. It is also shown that the uniqueness of pretangent spaces to subsets of a real line is closely related to the “asymptotic asymmetry” of these subsets. |
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Dovgoshey, O. Bilet, V. |
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Dovgoshey, O. Bilet, V. Uniqueness of spaces pretangent to metric spaces at infinity Український математичний вісник |
| author_facet |
Dovgoshey, O. Bilet, V. |
| author_sort |
Dovgoshey, O. |
| title |
Uniqueness of spaces pretangent to metric spaces at infinity |
| title_short |
Uniqueness of spaces pretangent to metric spaces at infinity |
| title_full |
Uniqueness of spaces pretangent to metric spaces at infinity |
| title_fullStr |
Uniqueness of spaces pretangent to metric spaces at infinity |
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Uniqueness of spaces pretangent to metric spaces at infinity |
| title_sort |
uniqueness of spaces pretangent to metric spaces at infinity |
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Інститут прикладної математики і механіки НАН України |
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2019 |
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https://nasplib.isofts.kiev.ua/handle/123456789/169432 |
| citation_txt |
Uniqueness of spaces pretangent to metric spaces at infinity / O. Dovgoshey, V. Bilet // Український математичний вісник. — 2019. — Т. 16, № 1. — С. 57-87. — Бібліогр.: 12 назв. — англ. |
| series |
Український математичний вісник |
| work_keys_str_mv |
AT dovgosheyo uniquenessofspacespretangenttometricspacesatinfinity AT biletv uniquenessofspacespretangenttometricspacesatinfinity |
| first_indexed |
2025-11-30T13:44:32Z |
| last_indexed |
2025-11-30T13:44:32Z |
| _version_ |
1850223124584333312 |
| fulltext |
Український математичний вiсник
Том 16 (2019), № 1, 57 – 87
Uniqueness of spaces pretangent
to metric spaces at infinity
Oleksiy Dovgoshey, Viktoriia Bilet
This paper dedicated to the memory of Professor Bogdan Bojarski
Abstract. We find the necessary and sufficient conditions under which
an unbounded metric space X has, at infinity, a unique pretangent space
ΩX
∞,r̃ for every scaling sequence r̃. In particular, it is proved that ΩX
∞,r̃
is unique and isometric to the closure of X for every logarithmic spiral
X and each r̃. It is also shown that the uniqueness of pretangent spaces
to subsets of real line is closely related to “asymptotic asymmetry” of
these subsets.
2010 MSC. 54E35.
Key words and phrases. Structure of metric space at infinity, loga-
rithmic spiral, asymmetric set of real numbers, rescaling.
1. Introduction
The present paper deals with the so-called pretangent spaces to metric
spaces at infinity introduced in [4, 5]. Papers [4, 5] as well as paper [6]
describe the structure of metric spaces with finite pretangent spaces at
infinity. The object of the present research is the metric spaces having,
for given scaling sequence, unique pretangent spaces at infinity.
The main results of the paper are the following.
• Theorem 2.1 giving the necessary and sufficient conditions under
which an unbounded metric space X has a unique pretangent space
ΩX∞,r̃ at infinity for every scaling sequence r̃.
• Theorem 3.1 which claims that each pretangent space to a logarith-
mic spiral at infinity is unique (for given scaling sequence), tangent,
and isometric to the closure of this spiral.
Received 12.03.2019
ISSN 1810 – 3200. c© Iнститут прикладної математики i механiки НАН України
58 Uniqueness of spaces pretangent to metric spaces...
• Theorem 3.2 describing the uniqueness of ΩX∞,r̃ for the unbounded
subgroups X of the multiplicative group of nonzero complex num-
bers.
• Theorem 4.1 characterizing the subsets X of R with unique ΩX∞,r̃
by symmetric properties of X.
The main results of the paper have natural infinitesimal analogs
(see [1, 2]).
Let us recall the definitions and denotations which will be used later.
Let r̃ = (rn)n∈N be a sequence of positive real numbers tending to
infinity, lim
n→∞
rn = ∞. In what follows r̃ will be called a scaling sequence
and, moreover, any formula of the form (xn)n∈N ⊂ A will mean that all
elements of the sequence (xn)n∈N belong to the set A.
Let (X, d) be an unbounded metric space.
Definition 1.1. Two sequences x̃ = (xn)n∈N ⊂ X and ỹ = (yn)n∈N ⊂ X
are mutually stable with respect to the scaling sequence r̃ if there is a
finite limit
lim
n→∞
d(xn, yn)
rn
:= d̃r̃(x̃, ỹ) = d̃(x̃, ỹ). (1.1)
Let p ∈ X. Denote by X̃∞ the set of all sequences x̃ = (xn)n∈N ⊂ X
for which limn→∞ d(xn, p) = ∞ and write Seq(X, r̃) for the set of all
sequences x̃ ∈ X̃∞ such that there is a finite limit
lim
n→∞
d(xn, p)
rn
:=
˜̃
dr̃(x̃).
Definition 1.2. A subset F of Seq(X, r̃) is self-stable if any two x̃, ỹ ∈ F
are mutually stable. F is maximal self-stable if it is self-stable and, for
arbitrary t̃ ∈ Seq(X, r̃), we have either t̃ ∈ F or there is x̃ ∈ F such that
x̃ and t̃ are not mutually stable,
lim inf
n→∞
d(xn, tn)
rn
< lim sup
n→∞
d(xn, tn)
rn
.
The maximal self-stable subsets of Seq(X, r̃) will be denoted by X̃∞,r̃.
Remark 1.1. The set X̃∞, the set Seq(X, r̃) and its self-stable and
maximal self-stable subsets are invariant under the choice of the point
p ∈ X. If x̃ ∈ Seq(X, r̃) and p, b ∈ X, then, using the triangle inequality,
we obtain
lim
n→∞
d(xn, p)
rn
= lim
n→∞
d(xn, b)
rn
.
O. Dovgoshey, V. Bilet 59
Let X̃∞,r̃ be a maximal self-stable subset of Seq(X, r̃). Consider the
function d̃ : X̃∞,r̃×X̃∞,r̃ → R defined by (1.1). Obviously, d̃ is symmetric
and nonnegative and d̃(x̃, x̃) = 0 holds for every x̃ ∈ X̃∞,r̃. Moreover,
the triangle inequality for d gives us the triangle inequality for d̃,
d̃(x̃, ỹ) ≤ d̃(x̃, z̃) + d̃(z̃, ỹ).
Hence (X̃∞,r̃, d̃) is a pseudometric space.
Now we are ready to define the main object of our research.
Definition 1.3. Let (X, d) be an unbounded metric space, let r̃ be a scal-
ing sequence and let X̃∞,r̃ be a maximal self-stable subset of Seq(X, r̃).
The pretangent space to (X, d) (at infinity, with respect to r̃) is the met-
ric identification of the pseudometric space (X̃∞,r̃, d̃).
Define a binary relation ≡ on X̃∞,r̃ as
(x̃ ≡ ỹ) ⇔
(
d̃r̃(x̃, ỹ) = 0
)
. (1.2)
It is easy to prove that ≡ is an equivalence relation on X̃∞,r̃. Write ΩX∞,r̃
for the set of equivalence classes generated by ≡ and define the function
ρ : ΩX∞,r̃ × ΩX∞,r̃ → R as
ρ(α, β) := d̃r̃(x̃, ỹ), x̃ ∈ α ∈ ΩX∞,r̃, ỹ ∈ β ∈ ΩX∞,r̃. (1.3)
Using Theorem 15 [9, Chapter 4] it can be proved that ρ is a well-defined
metric on ΩX∞,r̃. The metric identification of (X̃∞,r̃, d̃) is the metric space
(ΩX∞,r̃, ρ).
Let (nk)k∈N ⊂ N be a strictly increasing sequence. Denote by r̃′ the
subsequence (rnk
)k∈N of the scaling sequence r̃ = (rn)n∈N and, for every
x̃ = (xn)n∈N ∈ X̃∞, write x̃′ := (xnk
)k∈N. It is clear that we have
{x̃′ : x̃ ∈ Seq(X, r̃)} ⊆ Seq(X, r̃′)
and
˜̃
dr̃′(x̃
′) =
˜̃
dr̃(x̃) holds for every x̃ ∈ Seq(X, r̃). Furthermore, if se-
quences x̃, ỹ ∈ Seq(X, r̃) are mutually stable w.r.t. r̃, then x̃′ and ỹ′ are
mutually stable w.r.t. r̃′ and
d̃r̃(x̃, ỹ) = d̃r̃′(x̃
′, ỹ′). (1.4)
Consequently {x̃′ : x̃ ∈ X̃∞,r̃} is a self-stable subset of Seq(X, r̃′) for every
X̃∞,r̃. By Zorn’s lemma there is X̃∞,r̃′ such that
{x̃′ : x̃ ∈ X̃∞,r̃} ⊆ X̃∞,r̃′ ⊆ Seq(X, r̃′).
60 Uniqueness of spaces pretangent to metric spaces...
Let us denote by ϕr̃′ the mapping from X̃∞,r̃ to X̃∞,r̃′ with ϕr̃′(x̃) = x̃′.
It follows from (1.4) that, after corresponding metric identifications, the
mapping ϕr̃′ passes to an isometric embedding em′ : ΩX∞,r̃ → ΩX∞,r̃′ such
that the diagram
X̃∞,r̃
ϕr̃′−−−−→ X̃∞,r̃′
π
y
yπ′
ΩX∞,r̃
em′
−−−−→ ΩX∞,r̃′
(1.5)
is commutative. Here π and π′ are the natural projections,
π(x̃) = {ỹ ∈ X̃∞,r̃ : d̃r̃(x̃, ỹ) = 0} and π′(t̃) = {ỹ ∈ X̃∞,r̃′ : d̃r̃′(t̃, ỹ) = 0}.
Definition 1.4. Let (X, d) be an unbounded metric space and let r̃ be a
scaling sequence. A pretangent ΩX∞,r̃ is tangent if em′ : ΩX∞,r̃ → ΩX∞,r̃′ is
surjective for every ΩX∞,r̃′.
Remark 1.2. It can be proved that the following statements are equiv-
alent.
• The metric space ΩX∞,r̃ is tangent.
• The mapping em
′
: ΩX∞,r̃ → ΩX∞,r̃′ is an isometry for every ΩX∞,r̃′.
• The set {x̃′ : x̃ ∈ X̃∞,r̃} is a maximal self-stable subset of the set
Seq(X, r̃′) for every r̃′.
• The mapping ϕr̃′ : X̃∞,r̃ → X̃∞,r̃′ is onto for every X̃∞,r̃′ .
2. General criterion of uniqueness
We start from an example of a metric space having the unique pre-
tangent space for every scaling sequence.
Proposition 2.1. Let X = R+ = [0,∞) be the set of all non-negative
real numbers with the usual metric d(x, y) = |x− y| and let r̃ = (rn)rn∈N
be an arbitrary scaling sequence. Then the following statements hold for
every maximal self-stable set X̃∞,r̃.
(i) The equality X̃∞,r̃ = Seq(X, r̃) holds. In particular, a sequence
x̃ ∈ X̃∞ belongs to X̃∞,r̃ if and only if there is c ≥ 0 such that
˜̃dr̃(x̃) = lim
n→∞
xn
rn
= c. (2.1)
O. Dovgoshey, V. Bilet 61
(ii) Let x̃, ỹ ∈ X̃∞,r̃. Then the equality d̃r̃(x̃, ỹ) = 0 holds if and only
if
lim
n→∞
xn
rn
= lim
n→∞
yn
rn
.
(iii) The pretangent space ΩX∞,r̃ corresponding to X̃∞,r̃ is isometric to
(R+, |·, ·|) and tangent.
Proof. (i) The inclusion X̃∞,r̃ ⊆ Seq(X, r̃) is trivial. Suppose now that
x̃, ỹ ∈ Seq(X, r̃). Then choosing p = 0 in the definition of Seq(X, r̃) we
can find c1, c2 ∈ R+ such that
lim
n→∞
xn
rn
= c1, lim
n→∞
yn
rn
= c2.
Consequently,
lim
n→∞
|xn − yn|
rn
= |c1 − c2| (2.2)
holds, so that x̃ and ỹ are mutually stable. It implies statement (i).
(ii) Statement (ii) follows from statement (i) and (2.2).
(iii) Define a function f : ΩX∞,r̃ → R+ by the rule: if β ∈ ΩX∞,r̃
and x̃ ∈ β, then f(β) := limn→∞
xn
rn
. Statements (i), (ii) and limit
relation (2.2) imply that f is a well-defined isometry.
Let ñ = (nk)k∈N be a strictly increasing sequence of positive integer
numbers and let r̃′ = (rnk
)k∈N be the corresponding subsequence of the
scaling sequence r̃. Suppose x̃ ∈ X̃∞,r̃′ . Then there is b > 0 such that
lim
k→∞
xk
rnk
= b.
Define ỹ ⊂ X as
yn :=
xk if n = nk ∈ {n1, n2, . . .},
brn if n /∈ {n1, n2, . . .} and b > 0,
(rn)
1/2 if n /∈ {n1, n2, . . .} and b = 0.
It is clear that ỹ ∈ X̃∞ and ỹ′ = (ynk
)k∈N = x̃ and
lim
k→∞
yn
rn
= b.
By statement (i), ỹ belongs to X̃∞,r̃, i.e., Ω
X
∞,r̃ is tangent.
Statement (i) of Proposition 2.1 implies the following property of
the metric space (X, d) = (R, |·, ·|).
62 Uniqueness of spaces pretangent to metric spaces...
• For every scaling sequence r̃ there exists a unique pretangent space
ΩX∞,r̃.
We will denote by A the class of all unbounded metric spaces having
this property.
Remark 2.1. The uniqueness of pretangent spaces for (X, d) ∈ U is
understood in the usual set-theoretic sense. Each metric space is a pair
consisting of a set and a metric defined on the Cartesian square of this
set. The metric ρ defined on ΩX∞,r̃×ΩX∞,r̃ (see (1.1)) is evidently unique.
Thus the uniqueness of pretangent space (ΩX∞,r̃, ρ) means the uniqueness
of the set ΩX∞,r̃. The last set is the quotient set of X̃∞,r̃ generated by
≡ (see (1.2)). The equivalence relation ≡ on X̃∞,r̃ is unique for given
r̃. Consequently, ΩX∞,r̃ is unique if and only if X̃∞,r̃ is unique. Recall
that the construction of the maximal self-stable sets X̃∞,r̃ is based on
the choice of elements x̃ ∈ Seq(X, r̃). Thus the statement (X, d) ∈ A
claims that the pretangent spaces to (X, d) at infinity does not depend
on this choice.
Let (X, d) be a metric space and let p ∈ X. For each pair of nonempty
sets C, D ⊆ X, we write
∆(C,D) := sup{d(x, y) : x ∈ C, y ∈ D}.
In addition, for every ε ∈ (0, 1), we define the set S2
ε as
S2
ε :=
{
(r, t) ∈ Sp2(X) : r 6= 0 6= t and
∣∣∣r
t
− 1
∣∣∣ ≥ ε
}
, (2.3)
where Sp2(X) is the Cartesian square of Sp(X) = {d(x, p) : x ∈ X}.
Theorem 2.1. Let (X, d) be an unbounded metric space and let p be a
point of X. Then (X, d) ∈ A if and only if the following conditions are
satisfied simultaneously.
(i) The limit relations
lim
k→1
lim sup
r→∞
diam(A(p, r, k))
r
= lim
r→∞
diam(S(p, r))
r
= 0 (2.4)
hold, where r ∈ (0,∞), k ∈ [1,∞) and A(p, r, k) is the annulus
{x ∈ X :
r
k
≤ d(x, p) ≤ rk},
and S(p, r) is the sphere {x ∈ X : d(x, p) = r}.
O. Dovgoshey, V. Bilet 63
(ii) Let ε ∈ (0, 1). If ((qn, tn))n∈N ⊂ S2
ε and
lim
n→∞
qn = lim
n→∞
tn = ∞,
and there is
lim
n→∞
qn
tn
= c0 ∈ [0,∞], (2.5)
then there exists a finite limit
lim
n→∞
∆(S(p, qn), S(p, tn))
|qn − tn|
:= κ0. (2.6)
The following lemma is immediate from the definition of the pretan-
gent spaces.
Lemma 2.1. Let (X, d) be an unbounded metric space. Then the follow-
ing statements are equivalent.
(i) (X, d) ∈ A.
(ii) For every r̃, there is X̃∞,r̃ such that
X̃∞,r̃ = Seq(X, r̃). (2.7)
(iii) Equality (2.7) holds for every r̃ and every X̃∞,r̃.
(iv) All x̃, ỹ ∈ Seq(X, r̃) are mutually stable for every r̃.
Suppose (X, d) is a metric space. For all nonempty A, B ⊆ X we
define the distance from A to B as
inf{d(x, y) : x ∈ A, y ∈ B}
and denote it by δ(A,B). (See, for example, [11, Definition 2.7.1].)
Lemma 2.2. Let (X, d) be an unbounded metric space, p ∈ X. If con-
dition (i) from Theorem 2.1 holds, then we have the equality
lim
t,q→∞
(t,q)∈S2
ε
∆(S(p, q), S(p, t))
δ(S(p, q), S(p, t))
= 1 (2.8)
for every ε ∈ (0, 1).
64 Uniqueness of spaces pretangent to metric spaces...
Proof. Suppose there is ε ∈ (0, 1) such that (2.8) does not hold. Then
there exist two sequences (S(p, qn))n∈N and (S(p, tn))n∈N with
((tn, qn))n∈N ⊂ S2
ε
and four sequences x̃, ỹ, z̃, w̃ ∈ X̃∞ such that
xn, zn ∈ S(p, tn) and yn, wn ∈ S(p, qn),
for every n ∈ N, and
lim sup
n→∞
d(xn, yn)
d(zn, wn)
> 1. (2.9)
Suppose also that condition (i) of Theorem 2.1 holds. Then we obtain
lim
n→∞
d(xn, zn)
tn
= lim
n→∞
d(yn, wn)
qn
= 0. (2.10)
Note also that the relation (tn, qn) ∈ S2
ε and (2.3) imply that there is
ε1 ∈ (0, ε) such that the inequalities
|qn − tn| ≥ ε1qn and |tn − qn| ≥ ε1tn (2.11)
hold for every n ∈ N. Let us find the upper bound of the limits in (2.4).
Write
ηn :=
d(xn, zn)
tn
and ξn :=
d(yn, wn)
qn
for every n ∈ N. The triangle inequality implies
d(xn, yn)
d(zn, wn)
≤ d(xn, zn)
d(zn, wn)
+
d(zn, wn)
d(zn, wn)
+
d(wn, yn)
d(zn, wn)
≤ ηn
tn
d(zn, wn)
+ 1 + ξn
qn
d(zn, wn)
.
Using
d(zn, wn) ≥ |d(zn, p)− d(wn, p)| = |tn − qn|
and (2.11) we obtain
tn
d(zn, wn)
≤ tn
|tn − qn|
≤ 1
ε1
and
qn
d(zn, wn)
≤ qn
|tn − qn|
≤ 1
ε1
.
Thus,
lim sup
n→∞
d(xn, yn)
d(zn, wn)
≤ 1 +
1
ε1
lim sup
n→∞
(ηn + ξn).
The upper limit in the right is zero by (2.10). Consequently we have
lim sup
n→∞
d(xn, yn)
d(zn, wn)
≤ 1,
contrary to (2.9).
O. Dovgoshey, V. Bilet 65
Proof Theorem 2.1. Assume (X, d) ∈ A. We need verify conditions (i)
and (ii).
(i) Let us consider the function f : [1,∞) → R+ with
f(k) := k lim sup
r→∞
diam(A(p, r, k))
r
.
Since
f(k) = lim sup
r→∞
diam(A(p, k rk , k))
r
k
= lim sup
t→∞
diam(A(p, kt, k))
t
and
A(p, kt, k) = {x ∈ X : t ≤ d(x, p) ≤ k2t},
the function f is increasing. We evidently have
diam(A(p, r, k))
r
≤ 2rk
r
= 2k
for all k ≥ 1 and r > 0. Hence the double inequality
0 ≤ f(k) ≤ 2k2
holds. Consequently, there is a finite, positive limit
lim
k→1
f(k) := c0.
It is clear that this limit equals Ψ(1). Suppose that c0 > 0 and set
ε ∈ (0, c0). Then there is k0 > 1 such that the double inequality
c0 − ε < lim sup
r→∞
diam(A(p, r, k))
r
< c0 + ε (2.12)
holds for every k ∈ (1, k0]. Let (kn)n∈N ⊂ (1, k0] be a strictly decreasing
sequence such that
lim
n→∞
kn = 1. (2.13)
Double inequality (2.12) implies that there is a sequence r̃ = (rn)n∈N
such that limn→∞ rn = ∞ and
c0 − ε <
diam(A(p, rn, kn))
rn
< c0 + ε (2.14)
for every n ∈ N. It follows from (2.14) that there are x̃, ỹ ⊂ X such that
xn, yn ∈ A(p, rn, kn) and
d(xn, yn)
rn
≥ c0 − ε (2.15)
66 Uniqueness of spaces pretangent to metric spaces...
for every n ∈ N. The definition of A(p, rn, kn) and (2.15) imply that
d(xn, p)
rn
,
d(yn, p)
rn
∈
[
1
kn
, kn
]
(2.16)
for every n ∈ N. Define a sequence z̃ = (zn)n∈N ⊂ X by the rule
zn :=
{
xn if n is even,
yn if n is odd.
(2.17)
Then it follows from (2.13), (2.16) and (2.17) that
lim
n→∞
d(xn, p)
rn
= lim
n→∞
d(zn, p)
rn
= 1.
Moreover, (2.14) and (2.16) imply
lim inf
n→∞
d(xn, zn)
rn
= lim
n→∞
d(x2n, z2n)
r2n
= 0,
but
lim sup
n→∞
d(xn, zn)
rn
= lim sup
n→∞
d(x2n+1, z2n+1)
r2n+1
≥ c0 − ε > 0.
Thus x̃ and p̃ are mutually stable, z̃ and p̃ are mutually stable but x̃
and z̃ are not mutually stable (w.r.t. the scaling sequence r̃ = (rn)n∈N).
Hence, by Lemma 2.1, we have (X, d) /∈ A, contrary to the assumption.
(ii) Let ((qn, tn))n∈N be a sequence of elements of S2
ε such that
limn→∞ qn = limn→∞ tn = ∞ and (2.5) hold. If c0 = 0 or c0 = ∞
in (2.5), then it is clear that (2.6) holds with κ0 = 1, so it is sufficient to
set
c0 ∈ (0,∞). (2.18)
Let x̃, ỹ ∈ X̃∞ such that d(p, xn) = qn, d(p, yn) = tn for every n ∈ N and
lim
n→∞
d(xn, yn)
∆(S(p, qn), S(p, tn))
= 1. (2.19)
Let us consider the sequence q̃ = (qn)n∈N as a scaling sequence. Condi-
tions (2.5) and (2.18) imply that there is
˜̃
dq̃(ỹ) = lim
n→∞
d(yn, p)
qn
=
1
c0
<∞.
Hence, by Lemma 2.1, there is a finite limit
d̃q̃(x̃, ỹ) = lim
n→∞
d(xn, yn)
qn
.
O. Dovgoshey, V. Bilet 67
Moreover, since, for every n ∈ N, (qn, tn) ∈ S2
ε , we have c0 6= 1. Now
using (2.19) and (2.5) we obtain
lim
n→∞
∆(S(p, qn), S(p, tn))
|qn − tn|
= lim
n→∞
d(xn, yn)∆(S(p, qn), S(p, tn))
qn
∣∣∣1− tn
qn
∣∣∣ d(xn, yn)
= lim
n→∞
d(xn, yn)
qn
lim
n→∞
1∣∣∣1− tn
qn
∣∣∣
=
c0
|1− c0|
d̃q̃(x̃, ỹ). (2.20)
Condition (ii) is proved.
To prove that (i) and (ii) imply (X, d) ∈ A suppose (i) and (ii) are
satisfied simultaneously. We must show that ΩX∞,r̃ is unique for every
scaling sequence r̃. Let r̃ = (rn)n∈N be an arbitrary scaling sequence and
let x̃, ỹ ∈ Seq(X, r̃). By Lemma 2.1 it is sufficient to show that x̃ and ỹ
are mutually stable. If
˜̃
dr̃(x̃) = 0, then, by the triangle inequality,
lim sup
n→∞
d(xn, yn)
rn
≤ ˜̃
dr̃(x̃) +
˜̃
dr̃(ỹ) =
˜̃
dr̃(ỹ)
and
lim inf
n→∞
d(xn, yn)
rn
≥ ˜̃
dr̃(ỹ)− ˜̃
dr̃(x̃) =
˜̃
dr̃(ỹ).
Consequently
d̃r̃(x̃, ỹ) = lim
n→∞
d(xn, yn)
rn
=
˜̃
dr̃(ỹ),
holds. Thus x̃ and ỹ are mutually stable. The case
˜̃
dr̃(ỹ) = 0 is similar.
Hence, without loss of generality, we may assume that
˜̃dr̃(ỹ) 6= 0 6= ˜̃dr̃(x̃).
Consider first the case
˜̃
dr̃(ỹ) =
˜̃
dr̃(x̃) := b 6= 0.
This assumption implies that, for every k > 1, there is n0(k) ∈ N such
that the inclusion
A(p, brn, k) ⊇ {xn, yn} (2.21)
holds for all integer numbers n > n0(k). Recall that
A(p, brn, k) =
{
x ∈ X :
brn
k
≤ d(x, p) ≤ kbrn
}
.
68 Uniqueness of spaces pretangent to metric spaces...
It follows from (2.21) that
d(xn, yn) ≤ diam(A(p, brn, k))
if n > n0(k). Consequently,
1
b
lim sup
n→∞
d(xn, yn)
rn
≤ lim sup
n→∞
diam(A(p, brn, k))
brn
.
Letting k → 1 on the right-hand side of the last inequality and using (2.4)
we see that
0 ≤ 1
b
lim sup
n→∞
d(xn, yn)
rn
≤ lim
k→∞
(
lim sup
n→∞
diam(A(p, brn, k))
brn
)
= 0.
Hence
d̃r̃(x̃, ỹ) = lim
n→∞
d(xn, yn)
rn
= 0. (2.22)
It implies that x̃ and ỹ are mutually stable.
It still remains to show that there exists a finite limit
d̃r̃(x̃, ỹ) = lim
n→∞
d(xn, yn)
rn
if
0 6= ˜̃dr̃(x̃) 6= ˜̃dr̃(ỹ) 6= 0. (2.23)
For convenience we write
qn := d(xn, p), tn := d(yn, p)
for n ∈ N. Condition (2.23) implies that there are ε > 0 and an integer
number n0(ε) such that
max{qn, tn} > 0 and
∣∣∣∣
qn
tn
− 1
∣∣∣∣ ≥ ε (2.24)
for all n ≥ n0(ε). It is clear that
xn ∈ S(p, qn) and yn ∈ S(p, tn),
where S(p, qn) and S(p, tn) are the spheres with the common center p ∈ X
and the radii qn and tn, respectively. Consequently, we have the following
inequalities
∆(S(p, qn), S(p, tn)) ≥ d(xn, yn) ≥ δ(S(p, qn), S(p, tn)), (2.25)
O. Dovgoshey, V. Bilet 69
where the quantity δ(S(p, qn), S(p, tn)) is defined as in Lemma 2.2. Limit
relations (2.8) and (2.6) imply
κ0 = lim
n→∞
∆(S(p, qn), S(p, tn))
|qn − tn|
= lim
n→∞
δ(S(p, qn), S(p, tn))
|qn − tn|
.
Hence, using (2.25), we obtain
κ0 = lim
n→∞
d(xn, yn)
|qn − tn|
=
1∣∣∣ ˜̃dr̃(x̃)− ˜̃
dr̃(ỹ)
∣∣∣
lim
n→∞
d(xn, yn)
rn
,
that implies
d̃r̃(x̃, ỹ) = lim
n→∞
d(xn, yn)
rn
= κ0
∣∣∣ ˜̃dr̃(x̃)− ˜̃
dr̃(ỹ)
∣∣∣ . (2.26)
Thus x̃ and ỹ are mutually stable.
It can be proved that conditions (i) and (ii) from Theorem 2.1 are
mutually independent in the sense that none of them follows another.
We conclude this section by simple corollary of statement (iv) of
Lemma 2.1.
Proposition 2.2. Let (X, d) be a metric space and let Y be an unbounded
subspace X. Then (X, d) ∈ A implies (Y, d) ∈ A.
3. Uniqueness and logarithmic spirals
For simplicity we will consider only logarithmic spirals having the pole
at 0. The polar equation of these spirals is
ρ = kbϕ, (3.1)
where k and b are constants, k ∈ (0,∞) and b ∈ (0, 1) ∪ (1,∞). The
rotation of the polar axis on the angle ϕ1 = − ln k
ln b transforms (3.1) to the
form
ρ = bϕ. (3.2)
Let us denote by S∗ the set of all complex numbers lying on spiral (3.2)
and let
S = S
∗ ∪ {0},
i.e., S is the closure of S∗ in the complex plane C. In the following theorem
we consider S as a metric space with the usual metric d(z, w) = |z − w|.
Theorem 3.1. Each pretangent space to (S, d) at infinity is unique,
tangent and isometric to S.
70 Uniqueness of spaces pretangent to metric spaces...
We shall need the following lemma. Denote by C∗ the multiplicative
group of all nonzero complex numbers.
Lemma 3.1. S∗ is a subgroup of the group C∗.
Proof. As is well known, a non-void subset H of a group G is a subgroup
of G if and only if
hg ∈ H and h−1 ∈ H
for all h, g ∈ H. (See, for example, [10, Chapter 1, §2].) Let z be a point
of C∗. It is clear from (3.2) that z ∈ S∗ if and only if
z = |z| exp(i logb |z|). (3.3)
The last equality implies
z−1 = |z|−1 exp(−i logb |z|) =
∣∣z−1
∣∣ exp
(
i logb
∣∣z−1
∣∣) .
Hence z−1 belongs to S∗ if z ∈ S∗. Similarly we obtain zw ∈ S∗ if z ∈ S∗
and w ∈ S∗.
The next useful lemma describes the isometries of metric identifica-
tions of pseudometric spaces.
Lemma 3.2. Let (X, dX ) be a pseudometric space, (Y, dY ) a metric
space, (Ω, ρ) a metric identification of (X, dX ) and π : X → Ω the corre-
sponding natural projection. Then for every distance-preserving, surjec-
tive mapping F : X → Y there is a unique mapping f : Ω → Y such that
the diagram
X Y
Ω
F
π f
is commutative. The mapping f is an isometry of metric spaces (Ω, ρ)
and (Y, dY ).
Proof. Let us define a mapping f by the rule: if α ∈ Ω, then we set
f(α) := F (x), (3.4)
where x is an arbitrary point in π−1(α). This definition is correct. In-
deed, if x1, x2 ∈ π−1(α), then dX(x1, x2) = 0 because π is natural
projection. The equality dX(x1, x2) = 0 implies dY (F (x1), F (x2)) = 0
because F is distance-preserving. Since dY is a metric, the last equality
gives F (x1) = F (x2).
O. Dovgoshey, V. Bilet 71
Rewriting (3.4) as f(π(x)) = F (x) we see that the diagram is com-
mutative. The uniqueness of f which satisfies the equality F = f ◦ π
follows from the surjectivity of π. It still remains to prove that F is an
isometry.
Let α, β ∈ Ω, x, y ∈ X and α = π(x), β = π(y). Then we have
dY (f(α), f(β)) = dY (f(π(x)), f(π(y))) = dY (F (x), F (y))
= dX(x, y) = ρ(π(x), π(y)) = ρ(α, β).
Thus f is distance-preserving. Moreover f is surjective because F is
surjective. Hence f is an isometry as a distance-preserving, surjective
mapping between metric spaces.
Proof of Theorem 3.1. We shall first prove (S, d) ∈ U. To this end it
is sufficient to show that conditions (i) and (ii) from Theorem 2.1 are
satisfied with (X, d) = (S, d).
We start from the verification of condition (i).
Condition (i). Let z1 and z2 be some points of the annulus
A(0, r, k) =
{
z ∈ S :
r
k
≤ |z| ≤ kr
}
,
where k ∈ (1,∞), r ∈ (0,∞). Let us denote by l(z1, z2) the length of the
arc of spiral (3.2) joining the points z1, z2. If the polar coordinates of z1
and z2 are (ρ1, ϕ1) and (ρ2, ϕ2) respectively, then we obtain the famous
formula
l(z1, z2) =
∣∣∣∣
∫ ϕ2
ϕ1
√
ρ2 + ρ′2dϕ
∣∣∣∣ =
√
1 + ln2 b
| ln b| |bϕ2 − bϕ1 | (3.5)
=
√
1 + ln2 b
| ln b| |ρ2 − ρ1|.
It implies that
diamA(0, r, k) ≤ sup{l(z1, z2) : z1, z2 ∈ A(0, r, k)}
=
√
1 + ln2 b
| ln b|
∣∣∣∣rk − r
1
k
∣∣∣∣ =
r
√
1 + ln2 b
| ln b|
(
k − 1
k
)
.
Consequently,
lim
k→1
lim sup
r→0
diam(A(0, r, k))
r
≤ lim
k→1
√
1 + ln2 b
| ln b|
(
k − 1
k
)
= 0,
i.e., (2.4) holds and condition (i) is satisfied.
72 Uniqueness of spaces pretangent to metric spaces...
Condition (ii). Let ε ∈ (0, 1) and let ((qn, tn))n∈N be a sequence of
points of S2
ε such that
lim
n→∞
qn = lim
n→∞
tn = ∞ (3.6)
and there is
lim
n→∞
qn
tn
= c0 ∈ [0,∞]. (3.7)
We must show that there exists a finite limit
lim
n→∞
∆(S(0, qn), S(0, tn))
|qn − tn|
= κ0. (3.8)
It follows from the definition of the set S2
ε that qn, tn ∈ (0,∞) for all
n ∈ N. Consequently, we can find θn, τn ∈ (−∞,+∞) such that
qne
iτn ∈ S
∗ and tne
iθn ∈ S
∗ (3.9)
for every n ∈ N. Since the spheres S(0, qn) and S(0, tn) are single-point
and qne
iτn ∈ S(0, qn) and tne
iθn ∈ S(0, tn), we have
∆(S(0, tn), S(0, qn))
|tn − qn|
=
|tneiθn − qne
iτn |
|tn − qn|
with
τn = logb qn, θn = logb tn, (3.10)
(see formula (3.3)). Consider firstly equation (3.7) with c0 = 0 or c0 = ∞.
If c0 = 0, then we have
κ0 = lim
n→∞
|tneiθn − qne
iτn |
|tn − qn|
= lim
n→∞
|1− qn
tn
ei(τn−θn)|
|1− qn
tn
| =
|1− c0|
|1− c0|
= 1.
(3.11)
Similar computations yield κ0 = 1 for c0 = ∞. Suppose now c0 ∈ (0,∞).
Using (3.7) and (3.10) we obtain
lim
n→∞
(τn − θn) = lim
n→∞
(logb qn − logb tn) = lim
n→∞
logb
qn
tn
= logb c0. (3.12)
Moreover, we have
c0 6= 1 (3.13)
because ((qn, tn))n∈N ⊂ S2
ε . Applying (3.12) in (3.11) we obtain
lim
n→∞
|tneiθn − qne
iτn |
|tn − qn|
=
|1− c0 exp(i logb c0)|
|1− c0|
.
O. Dovgoshey, V. Bilet 73
Note that the right side in this equality is finite and well defined by virtue
of (3.13).
Thus conditions (i) and (ii) from Theorem 2.1 are satisfied, so that
this theorem implies the desirable uniqueness of pretangent spaces. It
still remains to prove that all pretangent spaces are tangent and isometric
to S.
Let r̃ = (rn)n∈N be a scaling sequence and let S̃∞,r̃ be the corre-
sponding maximal self-stable set. For every x̃ = (xn)n∈N ∈ S̃∞,r̃ there is
a unique x∗ ∈ S such that
|x∗| = lim
n→∞
d(xn, 0)
rn
=
˜̃
dr̃(x̃). (3.14)
We claim that the mapping
F : S̃∞,r̃ → S, F (x̃) = x∗ (3.15)
is surjective and distance-preserving in the sense that the equality
d̃r̃(x̃, ỹ) = |x∗ − y∗| (3.16)
holds for all x̃, ỹ ∈ S̃∞,r̃.
Surjectivity. We have already verified that (S, d) ∈ U. Hence, by
Lemma 2.1, it is sufficient to prove that for every a ∈ S there is x̃ ∈ S̃∞,r̃
such that |x∗| = |a|. If |a| = 0, then the equality |x∗| = |a| is evident
for x̃ ∈ S̃0∞,r̃. Assume that |a| > 0. For every n ∈ N define a complex
number z0n as
z0n :=
∣∣∣∣
1
rn
∣∣∣∣ exp
(
i log
∣∣∣∣
1
rn
∣∣∣∣
)
(3.17)
(see formula (3.3)). The points z0n and x∗ belong to S∗. By Lemma 3.1
the set S∗ is a subgroup of C∗. Since the equation cy = b is solvable in
every group with given b and c belonging to this group, there is xn ∈ S∗
such that
z0nxn = a. (3.18)
This equality and (3.17) imply
d(xn, 0)
rn
=
∣∣∣∣
xn
z0n
∣∣∣∣ = |a|
for every n ∈ N. Thus the equality |a| = |x∗| holds for x̃ := (xn)n∈N if
every xn fulfills (3.18).
Preservation of distances. Let x̃ and ỹ ∈ S̃∞,r̃. If x∗ = 0 or y∗ = 0,
then equality (3.16) follows simply from (3.14). Assume x∗ 6= 0 6= y∗.
This assumption implies x∗ ∈ S∗, y∗ ∈ S∗ and
xn ∈ S
∗, yn ∈ S
∗
74 Uniqueness of spaces pretangent to metric spaces...
for all sufficiently large n. These membership relations give, in particular,
the equalities
x∗ = |x∗| exp(i logb |x∗|) and xn = |xn| exp(i logb |xn|), (3.19)
see (3.3). Moreover, we can rewrite (3.17) as
z0n = |z0n| exp(i logb |z0n|).
Using it, (3.19) and (3.14) we obtain
lim
n→∞
|z0n||xn| = |x∗|. (3.20)
Consequently,
lim
n→∞
exp(i logb(|z0n||xn|)) = exp(i logb |x∗|). (3.21)
Relations (3.19)–(3.21) give the equality
lim
n→∞
z0nxn = x∗. (3.22)
Similarly, we have limn→∞ z0nyn = y∗. The last two limit relations imply
|x∗ − y∗| =
∣∣∣ lim
n→∞
(r0nxn − r0nyn)
∣∣∣ = lim
n→∞
|xn − yn|
rn
,
i.e., (3.16) holds.
We can now easily show that all pretangent spaces ΩS
∞,r̃ are isomet-
ric to S and tangent. Indeed, as has been shown above the mapping
F : S̃∞,r̃ → S is distance-preserving and onto. Hence, by Lemma 3.2,
there is an isometry f : ΩS
∞,r̃ → S. All ΩS
∞,r̃ are tangent if and only if
all mappings ϕr̃′ : S̃∞,r̃ → S̃∞,r̃′ are surjective (see diagram (1.5) and
Remark 1.2). Let A be an infinite subset of N, r̃′ = (rn)n∈A be the
subsequence of r̃ = (rn)n∈N and let ỹ = (yn)n∈A ∈ S̃∞,r̃′. Suppose that
˜̃dr̃′(ỹ) > 0 (the case ˜̃dr̃′(ỹ) = 0 is more simple and can be considered
similarly). Let us define xn as
xn :=
{
yn if n ∈ A
y
z0n
if n ∈ N \ A, (3.23)
where z0n is defined by (3.17) and y is a point of S∗ such that |y| = ˜̃
dr̃′(ỹ).
It follows from Lemma 3.1 that x̃ = (xn)n∈N ⊂ S. Moreover, (3.17) and
(3.23) imply
lim
n→∞
d(xn, 0)
rn
= |y| <∞. (3.24)
O. Dovgoshey, V. Bilet 75
Consequently, x̃ ∈ Seq(S, r̃) holds. It is also clear that x̃′ := (xn)n∈A = ỹ.
By Lemma 2.1, (S, d) ∈ U implies S̃∞,r̃ = Seq(S, r̃). Hence each ϕr̃′ is a
surjective mapping.
Remark 3.1. Lemma 3.1 is closely related to the contemporary defini-
tion of logarithmic spirals (see, for example, [3, 9.6.9.1]). In the proof
of Theorem 3.1 we have used formula (3.5) but we really need only in
the following remarkable fact discovered by Rene Descartes: The length
measured along the logarithmic spiral from the pole O to the point P of
the spiral is proportional to the radius vector OP .
The logarithmic spirals and the set R∗
+ := R+ \{0} of all strictly pos-
itive real numbers have some common properties: They are unbounded
subgroups of the multiplicative group C∗ and have unique pretangent
spaces at infinity. The following theorem shows that the logarithmic spi-
rals and R∗
+ are exhausted all maximal subgroups of C∗ having these
properties.
In what follows we denote by S∗(b) the set of complex numbers lying
on spiral (3.2).
Theorem 3.2. Let Γ∗ be an unbounded subgroup of the multiplicative
group C∗. The following two statements are equivalent.
(i) Γ∗ ⊆ R∗
+ or there is b ∈ (0, 1) ∪ (1,∞) such that Γ∗ ⊆ S∗(b).
(ii) Γ∗ ∈ U.
Proof. First of all we claim that the theorem is true for all unbounded
Γ∗ if and only if it is true for all closed (in C∗), unbounded Γ∗. Let
Γ
∗
be the closure of Γ∗ in C∗. For convenience we denote by (i) and
(ii) the statements obtained by the substitution of Γ
∗
for Γ∗ in (i) and,
respectively, in (ii). Since R∗
+ and S∗(b) are closed subsets of C∗ we have
the equivalences
(Γ
∗ ⊆ R
∗
+) ⇔ (Γ∗ ⊆ R
∗
+) and (Γ
∗ ⊆ S
∗(b)) ⇔ (Γ∗ ⊆ S
∗(b)).
Thus (i) ⇔ (i) holds. Using Lemma 3.2 we see that (ii) ⇔ (ii). More-
over, since the closure of a subgroup is a subgroup [10, p. 102], Γ
∗
is a
closed subgroup of C∗.
Thus we may assume, without loss of generality, that Γ∗ is a closed
subgroup of C∗.
(i) ⇒ (ii) This implication follows from Proposition 2.2, Theorem 3.1
and Proposition 2.1. Indeed, Proposition 2.1 claims, in particular, that
R+ has a unique pretangent space at infinity and, by Theorem 3.1, the
76 Uniqueness of spaces pretangent to metric spaces...
same property is true for S∗(b). Finally, the property “to have a unique
pretangent space” is hereditary by Proposition 2.2.
(ii) ⇒ (i) Suppose that Γ∗ is an unbounded, closed subgroup of C∗
and (ii) holds. Since Γ∗ is unbounded, there exists a sequence (pi)i∈N ⊂
Γ∗ with
lim
i→∞
|pi| = ∞.
Let z1, z2 ∈ Γ∗ and let |z1| = |z2| hold. Write ri =
|z1|
|pi| , i ∈ N. Note that
the points z1
pi
and z2
pi
belong to Γ∗ because Γ∗ is a subgroup of C∗. It is
clear that
zj
pi
∈ A(0, ri, k) = {z ∈ Γ∗ :
ri
k
≤ |z| ≤ kri}, j = 1, 2,
for all k ∈ [1,∞) and i ∈ N. Statement (ii) of the present theorem
implies (2.4). Consequently, by Theorem 2.1, we have
0 = lim
k→1
lim sup
i→∞
diam(A(0, ri, k))
ri
≥
∣∣∣z1pi −
z2
pi
∣∣∣
∣∣∣ z1pi
∣∣∣
=
|z1 − z2|
|z1|
,
i.e., the implication
(|z1| = |z2|) ⇒ (z1 = z2) (3.25)
is valid for all z1, z2 ∈ Γ∗.
Let us consider the continuous homomorphism Φ from the group Γ∗
to the group R∗
+ defined as
Φ(z) = |z|, z ∈ Γ∗.
Then Φ is closed. Consequently, Φ(Γ∗) is a closed subgroup of R∗
+. Using
the well-known classification of the closed subgroups of the additive group
for real numbers (see, for example, [7, Chapter V, §1, 1]), we obtain the
following three possibilities:
(i) Φ(Γ∗) = {1};
(ii) Φ(Γ∗) = R∗
+;
(iii) There is g ∈ (1,∞) such that Φ(Γ∗) = {gn : n ∈ Z}, where Z is the
set of integers.
Since Γ∗ is unbounded, the case (i1) is impossible. Implication (3.25)
shows that the homomorphism Φ : Γ∗ → R∗
+ is one-to-one. Consequently,
for the case (i3), the group Γ∗ is cyclic with the generator z = Φ−1(g).
O. Dovgoshey, V. Bilet 77
Writing z in the trigonometric form z = |z|eiϕ = geiϕ, ϕ ∈ [0, 2π), we
see that either Γ∗ ⊆ R∗
+ for ϕ = 0 or, for ϕ ∈ (0, 2π), Γ∗ lies on the
logarithmic spiral
S
∗(b) = {t exp(i logb t) : t ∈ R
∗
+}
with b = exp
(
ln g
ϕ
)
, where ln g is the natural logarithm of g.
Suppose now that we have Φ(Γ∗) = R∗
+. Then Φ is an isomorphism
of the groups Γ∗ and R∗
+ and, simultaneously, Φ is a homeomorphism as
a continuous, closed bijection of the topological spaces Γ∗ and R∗
+. Write
Φ−1 for the inverse mapping of Φ and
T := {z ∈ C : |z| = 1}.
Let Ψ: C∗ → T be the standard homomorphism, Ψ(z) = z
|z| and let
in : Γ∗ → C∗ be the inclusion, in(z) = z. Then the mapping
R
exp−→ R
∗
+
Φ−1
−→ Γ∗ in−→ C
∗ Ψ−→ T (3.26)
is a character (a continuous homomorphism) of the additive group R.
Denote this character by κ. Then there is ν ∈ R such that
κ(t) = exp(iνt) (3.27)
holds for all t ∈ R (see [10, p. 271]). Since Φ−1 is bijective, from (3.26)
and (3.27) we obtain that
κ(t) =
Φ−1(exp t)
exp t
= exp(iνt), t ∈ R,
i.e.,
Φ−1(|z|)
|z| = exp(iν ln |z|),
z = |z| exp(iν ln |z|) (3.28)
hold for every z ∈ Γ∗. The last equality implies Γ∗ ⊆ R∗
+ if ν = 0. For
ν 6= 0 we can rewrite (3.28) as
z = |z| exp(i logb |z|),
where b = exp
(
1
ν
)
. Hence Γ∗ is a logarithmic spiral with the pole at 0 if
ν 6= 0.
78 Uniqueness of spaces pretangent to metric spaces...
4. Uniqueness for subsets of R
In this section we specify the uniqueness conditions, presented by
Theorem 2.1 for the general, unbounded metric spaces X, to the case
X ⊆ R.
Lemma 4.1. Let X ⊆ R be unbounded and let d(x, y) = |x − y| for all
x, y ∈ X. Condition (i) from Theorem 2.1 does not hold if and only if
there exist some sequences x̃, ỹ ⊂ X such that
lim
n→∞
xn = lim
n→∞
yn = ∞ and xn ∈ (−∞, 0), yn ∈ (0,+∞) (4.1)
for every n ∈ N and
lim
n→∞
xn
yn
= −1. (4.2)
Proof. Without loss of generality we can suppose that 0 ∈ X.
Let x̃ and ỹ satisfy relations (4.1) and (4.2). For every n ∈ N, write
rn :=
√
|xn| |yn| and kn := max
{∣∣∣∣
xn
yn
∣∣∣∣
1
2
,
∣∣∣∣
yn
xn
∣∣∣∣
1
2
}
.
It is easy to see that
lim
n→∞
rn = ∞ and lim
n→∞
kn = 1,
and rn > 0, and kn ≥ 1, and
xn, yn ∈ A(0, rn, kn) = {x ∈ X :
rn
kn
≤ |x| ≤ rnkn}
for every n ∈ N. Now, using the inequality of arithmetic and geometric
means, we obtain
diam(A(0, rn, kn))
rn
≥ |yn − xn|
rn
=
|yn|+ |xn|
rn
=
√
|yn|
|xn|
+
√
|xn|
|yn|
≥ 2.
Hence the inequality
lim
k→1
lim sup
r→∞
A(0, r, k)
r
≥ 2
holds. Thus condition (i) of Theorem 2.1 does not hold.
Conversely, if condition (i) does not hold, then we have
lim
k→1
lim sup
r→∞
diam(A(0, r, k))
r
> 0. (4.3)
O. Dovgoshey, V. Bilet 79
We must find sequences x̃, ỹ ⊂ X which satisfy relations (4.1) and (4.2).
Inequality (4.3) implies that there are a constant c > 0 and some se-
quences (rn)n∈N ⊂ (0,∞) and (kn)n∈N ⊂ [1,∞) such that
lim
n→∞
rn = ∞ and lim
n→∞
kn = 1
and
diamA(0, rn, kn)
rn
> c (4.4)
for every n ∈ N. Let us consider the closed intervals
I+n :=
[
rn
kn
, rnkn
]
, I−n :=
[
−rnkn,−
rn
kn
]
.
It is clear that
A(0, rn, kn) ⊆ I+n ∪ I−n .
Inequality (4.4) implies, for every n ∈ N, there are xn, yn ∈ A(0, rn, kn)
such that xn < yn and
|xn − yn|
rn
> c. (4.5)
If xn, yn ∈ I+n or xn, yn ∈ I−n , then
|xn − yn|
rn
≤ kn −
1
kn
. (4.6)
Since limn→∞ kn = 1, inequality (4.6) contradicts (4.5) for sufficiently
large n. Hence
xn ∈ I−n and yn ∈ I+n (4.7)
if n is large enough. Relations (4.7) and limn→∞ kn = 1 imply (4.2). The
rest of desirable properties of x̃ = (xn)n∈N and ỹ = (yn)n∈N are evident
by construction.
Let X ⊆ R and p ∈ R. Recall that X is locally symmetric at a point
p ∈ R if there is ε > 0 such that
(x+ p ∈ X) ⇔ (−x+ p ∈ X)
holds for every x ∈ X ∩ (p− ε, p+ ε). (See, for example, [12, p. 225].)
We shall say a set X ⊆ R is asymmetric at infinity with respect to a
point p ∈ R if there is ε > 0 such that
(x+ p ∈ X) ⇒ (−x+ p /∈ X) (4.8)
for each x ∈ X ∩
(
(−∞, p − ε) ∪ (p + ε,∞)
)
.
80 Uniqueness of spaces pretangent to metric spaces...
Corollary 4.1. Let X ⊆ R be unbounded. If condition (i) from The-
orem 2.1 holds, then X is asymmetric at infinity with respect to every
point p ∈ R.
Example 4.1 (Asymmetric at Infinity Subset of R). Let (rn), n ∈ N ∪
{0}, be a strictly convex sequence with r1 > r0 = 0 and
lim
n→∞
(rn+1 − rn) = ∞. (4.9)
Recall that the strict convexity of (rn) means that the second order
differences
∆2rn = rn+2 − 2rn+1 − rn
are strictly positive for every n ≥ 0. It implies
rn+2 − rn+1 > rn+1 − rn > . . . > r1 − r0 = r1 > 0. (4.10)
Write
X+ :=
∞⋃
n=1
[r4n, r4n+1], X− :=
∞⋃
n=1
[−r4n−1,−r4n−2]
and define
X := X+ ∪X−.
It is clear that X is an unbounded subset of R. We claim that X is
asymmetric at infinity with respect all points p ∈ R.
Let p ∈ R be given. We must show that −x+p /∈ X if x+p ∈ X and
|x| is large enough. Using (4.9) and (4.10) we can find n0 ∈ N such that
2|p| < rn0 − rn0−1. (4.11)
If x+ p ∈ X, then there is m ∈ N ∪ {0} such that either
x+ p > 0 and x+ p ∈ [r4m, r4m+1] (4.12)
or
x+ p < 0 and x+ p ∈ [−r4m−1,−r4m−2]. (4.13)
Suppose (4.12) holds (case (4.13) is similar). For sufficiently large x we
obtain the inequality 4m > n0. It follows from (4.12) that
−x+ p ∈ [−r4m+1 + 2p,−r4m + 2p] ⊆ [−r4m+1 − 2|p|,−r4m + 2|p|].
Now, using the inequality 4m > n0, (4.12), (4.10) we obtain
−x+ p ∈ [−r4m+1 − |rn0 − rn0−1|,−r4m + |rn0 − rn0−1|]
⊂ (−r4m+2,−r4m−1).
It follows from the definition of the set X, that the intersection of X
with the interval (−r4m+2,−r4m−1) is empty. Thus we have −x+p /∈ X.
O. Dovgoshey, V. Bilet 81
Consider now the “real-valued” variant of condition (ii) from Theo-
rem 2.1.
Let X ⊆ R be asymmetric at infinity with respect to p ∈ X. Write
X+
p := {x ∈ X : x+ p ∈ X, x+ p ≥ 0},
X−
p := {x ∈ X : x+ p ∈ X, x+ p ≤ 0}.
Define the subsets +1Rp,X and −1Rp,X of the set R+ by the rules:
+1Rp,X := {|x− p| : x ∈ X+
p } and −1Rp,X := {|x− p| : x ∈ X−
p }.
Then we obtain
Sp(X) = +1Rp,X ∪ −1Rp,X and +1Rp,X ∩ −1Rp,X ∩ [δ,∞) = ∅,
for sufficiently large δ > 0. Simple geometric considerations show that,
for all q, t ∈ Sp(X) ∩ [δ,∞), we have
∆(S(p, q), S(p, t)) =
{
|q − t| if (q, t) ∈ (+1R
2
p,X) ∪ (−1R
2
p,X),
|q + t| otherwise,
(4.14)
where +1R
2
p,X and −1R
2
p,X are the Cartesian squares of +1Rp,X and, re-
spectively, of −1Rp,X and
∆(S(p, q), S(p, t)) = sup{|x− y| : x ∈ S(p, q), y ∈ S(p, t)}.
For every δ > 0, let us introduce also the sets
+Kδ :=
{q
t
: q, t ∈ +1Rp,X ∩ [δ,∞)
}
, (4.15)
−Kδ :=
{q
t
: q, t ∈ −1Rp,X ∩ [δ,∞)
}
,
1Kδ :=
{q
t
: (q, t) ∈ (Sp2(X) ∩ [δ,∞)2) \ (+1R
2
p,X ∪ −1R
2
p,X)
}
,
where [δ,∞)2 and Sp2(X) are the Cartesian squares of the infinite interval
[δ,∞) and Sp(X).
Proposition 4.1. Let X ⊆ R be unbounded and asymmetric at infinity
with respect to p ∈ X. Condition (ii) of Theorem 2.1 holds if and only if
⋂
δ∈R∗
+
(Cl(+Kδ ∪ −Kδ) ∩ Cl(1Kδ)) ⊆ {0, 1,∞}, (4.16)
where R∗
+ = (0,∞) and the closures are taken in [0,∞].
82 Uniqueness of spaces pretangent to metric spaces...
Proof. Suppose inclusion (4.16) holds. Let ε0 > 0 and let ((qn, tn))n∈N ⊂
S2
ε0 such that limn→∞ qn = limn→∞ tn = ∞ and
lim
n→∞
qn
tn
:= c0 ∈ [0,∞]. (4.17)
It is necessary to show that there is a finite limit
lim
n→∞
∆(S(p, qn), S(p, tn))
|qn − tn|
= κ0. (4.18)
We first note that (4.18) holds with κ0 = 1 if c0 = 0 or c0 = ∞. Indeed,
equality (4.14) implies the double estimation
|qn − tn|
|qn − tn|
≤ ∆(S(p, qn), S(p, tn))
|qn − tn|
≤ |qn + tn|
|qn − tn|
.
Letting n to infinity and using (4.17) with c0 ∈ {0,∞} we obtain (4.18)
with κ0 = 1. Let us consider now the case 0 < c0 < ∞. Define, for
δ > 0,
Kδ :=
{q
t
: q, t ∈ Sp(X) ∩ [δ,∞)
}
.
Then, using the standard inclusion from the theory of cluster sets (see,
for example, [8, 1.1]), we have
c0 ∈
⋂
δ∈R∗
+
Cl(Kδ).
Furthermore, it follows from (4.15) that
Kδ = 1Kδ ∪ (+Kδ ∪ −Kδ).
Thus Cl(Kδ) = Cl(1Kδ) ∪ Cl(+Kδ ∪ −Kδ). The last equality and the
monotonicity: “if δ1 ≥ δ2, then
Cl(1Kδ1) ⊆ Cl(1Kδ2) and Cl(+Kδ1 ∪ −Kδ1) ⊆ Cl(+Kδ2 ∪ −Kδ2), ”
(4.19)
imply the equality
⋂
δ∈R∗
+
Cl(Kδ) =
( ⋂
δ∈R∗
+
Cl(1Kδ)
)⋃( ⋂
δ∈R∗
+
Cl(+Kδ ∪ −Kδ)
)
. (4.20)
Hence we have
c0 ∈
⋂
δ∈R∗
+
Cl(1Kδ) (4.21)
O. Dovgoshey, V. Bilet 83
or
c0 ∈
⋂
δ∈R∗
+
Cl(+Kδ ∪ −Kδ). (4.22)
It follows directly form (4.16) that
( ⋂
δ∈R∗
+
Cl(1Kδ)
)⋂( ⋂
δ∈R∗
+
Cl(+Kδ ∪ −Kδ)
)
⊆ {0, 1,∞}.
The relation (qn, tn) ∈ S2
ε0 imply the inequality
∣∣∣∣
qn
tn
− 1
∣∣∣∣ ≥ ε0,
so that
c0 6= 1.
Since c0 /∈ {0, 1,∞}, (4.21) and (4.22) imply that there is n0 ∈ N such
that either
(qn, tn) ∈ Sp2(X) \ (+1R
2
p,X ∪ −1R
2
p,X) (4.23)
for every n ≥ n0, or
(qn, tn) ∈ +1R
2
p,X ∪ −1R
2
p,X (4.24)
for every n ≥ n0. Now applying (4.14) we obtain
κ0 =
{ |1+c0|
|1−c0| if (4.23) holds,
1 if (4.24) holds.
The “sufficiency” is proved.
To prove the “necessity” suppose that (4.16) does not hold. The left
side of (4.16) can be written as
( ⋂
δ∈R∗
+
Cl(+Kδ ∪ −Kδ)
)⋂( ⋂
δ∈R∗
+
Cl(1Kδ)
)
.
Consequently there is c0 ∈ (0,∞) such that c0 6= 1 and
c0 ∈
⋂
δ∈R∗
+
Cl(+Kδ ∪ −Kδ) and c0 ∈
⋂
δ∈R∗
+
Cl(1Kδ).
Hence there are sequences
((zn, wn))n∈N ⊂ (+1R
2
p,X ∪ −1R
2
p,X) (4.25)
84 Uniqueness of spaces pretangent to metric spaces...
and
((qn, tn))n∈N ⊂ (Sp2(X)) \ (+1R
2
p,X ∪ −1R
2
p,X) (4.26)
such that
lim
n→∞
qn = lim
n→∞
tn = lim
n→∞
zn = lim
n→∞
wn = ∞ (4.27)
and
lim
n→∞
qn
tn
= lim
n→∞
zn
wn
= c0. (4.28)
Since c0 6= 1, there is ε0 > 0 such that the inequalities
∣∣∣∣
qn
tn
− 1
∣∣∣∣ ≥ ε0 and
∣∣∣∣
zn
wn
− 1
∣∣∣∣ ≥ ε0 (4.29)
hold if n is sufficiently large. Using (4.14) and (4.25), (4.26) we obtain
lim
n→∞
∆(S(p, qn), S(p, tn))
|qn − tn|
=
1 + c0
|1− c0|
(4.30)
and
lim
n→∞
∆(S(p, zn), S(p,wn))
|zn − wn|
= 1. (4.31)
Let n0 ∈ N be a number such that (4.29) holds for every n ≥ n0. Define
a sequence ((sn, yn))n∈N as a “mixture” of the sequences ((qn, tn))n∈N
and ((zn, wn))n∈N,
(sn, yn) =
(qn0 , tn0) if n ≤ n0
(qn, tn) if n is add and n > n0
(zn, wn) if n is even and n > n0.
Then ((sn, yn))n∈N ⊂ S2
ε0 and, by (4.27), (4.28), we have
lim
n→∞
sn = lim
n→∞
yn = ∞, lim
n→∞
sn
yn
= c0.
If there is
lim
n→∞
∆(S(p, sn), S(p, yn))
|sn − yn|
,
then the definition of the sequence ((sn, yn))n∈N and equalities (4.30),
(4.31) imply
1 =
1 + c0
|1− c0|
.
Hence c0 = 0, contrary to the condition c0 /∈ {0, 1,∞}.
Thus condition (ii) from Theorem 2.1 is false if (4.16) is false.
O. Dovgoshey, V. Bilet 85
The following theorem is the main result of the present section. De-
fine, for δ > 0,
+−Kδ :=
{q
t
: q ∈ +1Rp,X ∩ [δ,∞), t ∈ −1Rp,X ∩ [δ,∞)
}
, (4.32)
where q
t := ∞ if −1Rp,X ∩ [δ,∞) = ∅ and q
t := 0 if +1Rp,X ∩ [δ,∞) = ∅.
Theorem 4.1. Let X be an infinite subset of R, let d(x, y) = |x− y| for
all x, y ∈ X and let p be a point of X. Then (X, d) ∈ U if and only if X
is asymmetric at infinity with respect to p and
⋂
δ∈R∗
+
(Cl(+Kδ ∪ −Kδ) ∩ Cl(+−Kδ)) ⊆ {0,∞}. (4.33)
Proof. Suppose (X, d) ∈ U. Then conditions (i) and (ii) from Theo-
rem 2.1 holds. Using Corollary 4.1 we see thatX is asymmetric at infinity
w.r.t. p. Moreover, since 1Kδ ⊇ +−Kδ, we obtain from inclusion (4.16)
the inclusion
⋂
δ∈R∗
+
(Cl(+Kδ ∪ −Kδ) ∩ Cl(+−Kδ)) ⊆ {0, 1,∞}.
It still remains to prove that
1 /∈
⋂
δ∈R∗
+
(Cl(+Kδ ∪ −Kδ) ∩ Cl(+−Kδ)). (4.34)
Statement (4.34) holds if
1 /∈
⋂
δ∈R∗
+
Cl(+−Kδ). (4.35)
Suppose (4.35) does not hold. Then there exist some sequences
(xn)n∈N ⊂ (−∞, p] ∩X and (yn)n∈N ⊂ [p,∞) ∩X
such that
lim
n→∞
|xn − p| = lim
n→∞
|yn − p| = ∞
and
lim
n→∞
p− xn
yn − p
= 1. (4.36)
By Lemma 4.1 it contradicts condition (i) from Theorem 2.1. Thus (4.34)
is proved.
86 Uniqueness of spaces pretangent to metric spaces...
Conversely, assume inclusion (4.33) holds and X is asymmetric at
infinity w.r.t. p. We must prove conditions (i) and (ii) from Theorem 2.1.
Rewriting Lemma 4.1 in the terms of cluster sets we see that condi-
tion (i) does not hold if and only if
1 ∈
⋂
δ∈R∗
+
Cl(+−Kδ).
This relation and
1 ∈
⋂
δ∈R∗
+
Cl(+Kδ ∪ −Kδ)
show that the point 1 belongs to the set in the left part of formula (4.33),
contrary to the supposition. Condition (i) follows.
To prove condition (ii) from Theorem 2.1 note that
1Kδ = +−Kδ ∪ −+Kδ
for sufficiently large δ > 0 because X is asymmetric at infinity w.r.t.
p, where 1Kδ is defined by (4.15) and −+Kδ is the set obtained by the
permutation of the symbols “+” and “−” in (4.32). Consequently, by
Proposition 4.1, condition (ii) holds if
⋂
δ∈R∗
+
(Cl(+Kδ ∪ −Kδ) ∩ Cl(+−Kδ ∪ −+Kδ)) ⊆ {0, 1,∞}. (4.37)
Similarly (4.20) we can show that
⋂
δ∈R∗
+
(Cl(+Kδ ∪ −Kδ) ∩ Cl(+−Kδ ∪ −+Kδ))
=
( ⋂
δ∈R∗
+
(
Cl(+Kδ ∪ −Kδ) ∩ Cl(+−Kδ)
))
∪
( ⋂
δ∈R∗
+
(
Cl(+Kδ ∪ −Kδ) ∩ Cl(−+Kδ)
))
. (4.38)
It follows from the definition of the sets +Kδ, −Kδ, +−Kδ and −+Kδ that
if a positive number s belongs to
⋂
δ∈R∗
+
(Cl(+Kδ ∪ −Kδ) ∩Cl(−+Kδ)),
then
1
s
∈
⋂
δ∈R∗
+
(Cl(+Kδ ∪ −Kδ) ∩ Cl(+−Kδ)).
O. Dovgoshey, V. Bilet 87
Consequently, we can permute “+” and “−” (4.33). Thus we have the
inclusion
⋂
δ∈R∗
+
(Cl(+Kδ ∪ −Kδ) ∩ Cl(+−Kδ ∪ −+Kδ)) ⊆ {0,∞},
which implies (4.37).
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[12] B. S. Thomson, Symmetric Properties of Real Functions, Marcel Dekker Inc.,
New York, 1994.
Contact information
Oleksiy Dovgoshey Institute of Applied Mathematics
and Mechanics of the NASU,
Slov’yansk, Ukraine
E-Mail: oleksiy.dovgoshey@gmail.com
Viktoriia Bilet Institute of Applied Mathematics
and Mechanics of the NASU,
Slov’yansk, Ukraine
E-Mail: viktoriiabilet@gmail.com
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